Step 1: Apply the principle of conservation of mechanical energy. Since the system is released from rest and the pulley is frictionless, mechanical energy is conserved. The initial potential energy of $M_2$ is converted into kinetic energy of both masses and potential energy gained by $M_1$. The initial total energy ($E_i$) is the potential energy of $M_2$ (assuming $M_1$ starts at a reference height of $0$). $$E_i = M_2 g h_2$$ The final total energy ($E_f$) is the potential energy of $M_1$ (which has moved up by $h_2$) plus the kinetic energy of both masses just before $M_2$ hits the ground. $$E_f = M_1 g h_2 + \frac{1}{2} M_1 v_f^2 + \frac{1}{2} M_2 v_f^2$$ By conservation of energy, $E_i = E_f$. $$M_2 g h_2 = M_1 g h_2 + \frac{1}{2} M_1 v_f^2 + \frac{1}{2} M_2 v_f^2$$ Step 2: Rearrange the equation to solve for the final velocity $v_f$. $$M_2 g h_2 - M_1 g h_2 = \frac{1}{2} (M_1 + M_2) v_f^2$$ $$g h_2 (M_2 - M_1) = \frac{1}{2} (M_1 + M_2) v_f^2$$ $$v_f^2 = \frac{2 g h_2 (M_2 - M_1)}{M_1 + M_2}$$ $$v_f = \sqrt{\frac{2 g h_2 (M_2 - M_1)}{M_1 + M_2}}$$ Step 3: Substitute the given values into the equation. Given: $M_1 = 1.5 \text{ kg}$ $M_2 = 4.5 \text{ kg}$ $h_2 = 2 \text{ m}$ $g = 9.81 \text{ m/s}^2$ $$v_f = \sqrt{\frac{2 \times 9.81 \text{ m/s}^2 \times 2 \text{ m} \times (4.5 \text{ kg} - 1.5 \text{ kg})}{1.5 \text{ kg} + 4.5 \text{ kg}}}$$ $$v_f = \sqrt{\frac{2 \times 9.81 \times 2 \times (3.0)}{6.0}} \text{ m/s}$$ $$v_f = \sqrt{\frac{117.72}{6.0}} \text{ m/s}$$ $$v_f = \sqrt{19.62} \text{ m/s}$$ $$v_f \approx 4.429 \text{ m/s}$$ The velocity with which $M_2$ hits the ground is approximately $4.43 \text{ m/s}$. $$\boxed{\text{4.43 m/s}}$$ That's 2 down. 3 left today — send the next one.